JEE Mains · Physics · STD 12 - 2. Electric potential and capacitance
A capacitor of capacitance \(C\) is charged to a potential \(V\). The flux of the electric field through a closed surface enclosing the positive plate of the capacitor is \(......\)
- A \(\frac{ CV }{2 \varepsilon_0}\)
- B \(\frac{2 CV }{\varepsilon_0}\)
- C \(\frac{ CV }{\varepsilon_0}\)
- D Zero
Answer & Solution
Correct Answer
(C) \(\frac{ CV }{\varepsilon_0}\)
Step-by-step Solution
Detailed explanation
\(\phi=\frac{q_{\text {in }}}{\epsilon_0}\) \(=\frac{Q}{\epsilon_0}\) \(=\frac{C V}{\epsilon_0}\)
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